Angle Bisectors in Ellipse
Let A and B be two points on an ellipse with foci E and F. The tangents to the ellipse at A and B meet in S. Prove that ∠AES = ∠SEB. In words, the point of intersection of two tangents to an ellipse lies on the bisector of the angle formed by joining a focus to the points of tangency.
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Proof
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Copyright © 1996-2012Alexander Bogomolny
Proof
Reflect F in the two tangents to obtain points M and N. Because of the reflective property of ellipse, B is collinear with E, N and A with E, M.
From the definition of ellipse, EA + FA = EB + FB. Since M and N are reflections of F, FB = BN and FA = AM, implying
| | EM | = EA + AM |
| | | = EA + FA |
| | | = EB + FB |
| | | = EB + BN |
| | | = EN. |
It follows that ΔMEN is isosceles with EM = EN.
On the other hand, in ΔMFN, tangents AS and BS are the perpendicular bisectors of sides FN and FM. They meet in the circumcenter of ΔMFN, which also lies on the perpendicular bisector of the third side MN. By the construction, S is the point of intersection of the two tangents. Thus S is also the circumcenter of ΔMFN and lies on the perpendicular bisector of MN.
Now return to ΔMEN. Since it is isosceles with base MN, the perpendicular bisector of MN passes through E and plays the role of the bisector of ∠MEN which thus passes through S. But MEN and AEB is one and the same angle, and we are done.
(I thank Hubert Shutrick for his selfless help.)
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Copyright © 1996-2012Alexander Bogomolny
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